# Probability of passing a T/F exam?

There are $$n$$ questions in an examination ($$n \in \mathbb{N}$$), and the answer to each question is True or False. You know that exactly $$t$$ of the answers are True ($$0 \le t \le n$$), so you randomly answer $$t$$ questions as True and the rest as False. What is your probability of getting at least a $$50\%$$ score in the exam (in terms of $$n$$ and $$t$$)?

My attempt:

WLOG assume you answer the first $$t$$ questions as True. Let there be $$k$$ questions out of the first $$t$$ of which the answer is True. Thus, out of the other $$n-t$$ questions where you replied False, $$(n-t)-(t-k)=n-2t+k$$ questions are really False. Thus, the fraction of correct answers for the whole examination is equal to $$\frac{n-2t+2k}{n}$$. If this is at least $$1/2$$, then $$n+2k \ge 4t$$. However, I can't calculate the probability of this happening.

• A suggestion: Even-case: The probability to answer at least $t/2$ of the answers right is $\sum\limits_{k=\frac{t}{2}}^n \binom{n}{k}\cdot \left(\frac12\right)^n$ Dec 3, 2021 at 14:45
• If there are $10$ questions and $9$ of them are true and $1$ false and you answer $9$ as true, you are certain to get more than $50 \%$ score. So $t / n$ matters. Dec 3, 2021 at 14:52

There is a typo. Instead of $$n+2k \ge 4t$$, it should be $$n+4k \ge 4t$$. I do not know if you can get a closed form but here is my work that may help you understand lower and upper bound of $$t$$ in terms of $$n$$, beyond which one is certain to get at least $$50\%$$ score.

a) Based on your work, $$\displaystyle k \geq \frac{4t - n}{4}$$. So if $$t \geq \dfrac{3n}{4}, k \geq \dfrac{n}{2}$$ and we are bound to get at least $$50\%$$ score by choosing $$t$$ answers as TRUE.

Similarly if $$~\displaystyle t \leq \frac{n}{4}$$, you are certain to get at least $$50\%$$ score by randomly choosing $$(n-t)$$ answers as FALSE.

b) Now for $$~ \displaystyle \frac n4 \lt t \lt \frac {3n}4$$,

we must have $$\displaystyle \lceil \frac{4t - n}{4} \rceil \leq k \leq t$$ to score at least $$50\%$$

So the desired probability can be written as,

$$\displaystyle \sum \limits_{k = k_l}^t {t \choose k} {n - t \choose t - k} / {n \choose t}$$

where $$\displaystyle k_l = \lceil \frac{4t - n}{4} \rceil$$